Area Under the Curve

The area under the curve is a concept in integral calculus that quantifies the total area enclosed by a curve, the x-axis, and specified vertical boundaries on the graph of a function. This area is calculated using definite integrals and represents the accumulation of a quantity described by the function over a given interval.

Mathematical Definition

For a continuous function f(x) defined on the interval [a, b], the area under the curve from x = a to x = b is given by the definite integral: $\text{ Area }={\int }_a^{b}f(x)dx$

1.0Area Under the Curve Definition

The area under the curve is a fundamental concept in integral calculus that quantifies the total area between the graph of a function and the x-axis over a specified interval. Mathematically, it is determined using definite integrals.

Mathematical Representation:

If f(x) is a continuous function defined on the interval [a, b], the area under the curve from x = a to x = b is given by the definite integral: $\text{ Area }={\int }_a^{b}f(x)dx$

2.0Area Under the Curve – Between a Curve and Coordinate Axis

Area under the Curve and x- axis (Vertical Strips)

a.

Area of the strip = y.dx

Area bounded by the curve, the x-axis and the ordinate at x = a and x = b is given by A=${\int }_a^{b}ydx\text{, where }y=f(x)$ lies above the x-axis and b > a. 

Here vertical strip of thickness dx is considered at distance x.

b.

If y =  f(x)  lies completely below the x-axis, then  $A=\left|{\int }_a^{b}ydx\right|$

c.

If curve crosses the $x_{\text{-axis at }}x=c\text{, then }A=\left|{\int }_a^{c}ydx\right|+{\int }_c^{b}ydx$ 

Area under the Curve and y- axis (Horizontal Strips)

a.

Area of the strip = x.dy

Graph of x = g(y)

If g(y) $\ge 0\text{ for }y\in [c,d]$ then area bounded by curve x = g(y) and y-axis between abscissa y = c and y = d is ${\int }_{y=c}^{d}g(y)dy$

b. If g(y) $\le 0\text{ for }y\in [c,d]$ then area bounded by curve x = g(y) and y-axis between abscissa y = c and y = d is -${\int }_{y=c}^{d}g(y)dy$

Note:

General formula for area bounded by curve x = g(y) and y-axis between abscissa y = c and y=d $\text{ is }{\int }_{y=c}^d|g(y)|dy$.

3.0Area under the curve - Symmetric Area

If the curve is symmetric in all four quadrants, then 

Total area = 4 (Area in any one of the quadrants).

4.0Area Under a curve - Between Two Curves       

1. Area bounded by two curves y=f(x) \& y=g(x) 

such that f(x)>g(x) is 

 $A={\int }_{x_1}^{x_2}[f(x)-g(x)]dx$ where  x₁ and x₂ are roots of equation f(x) = g (x) 

2. In case horizontal strip is taken we have

$A={\int }_{y_1}^{y_2}[f(y)-g(y)]dy$

Where y₁ & y₂ are roots of equation f(y) = g(y) 

3. If the curves y = f(x) and y = g(x) intersect at x = c , then required area $A={\int }_a^c(g(x)-f(x))dx+{\int }_c^b(f(x)-g(x))dx={\int }_a^b|f(x)-g(x)|dx$

Note: Required area must have all the boundaries indicated in the problem.

5.0Standard Areas (To be Remembered)

1. Area bounded by parabolas $y^2=4ax;x^2=4by,a>0;b>0is\frac{16ab}{3}$ 

Intersection point:

$\begin{array}{l}{x}^2=4by\\ y^2=4ax\end{array}\}$

$x^2=4b\sqrt{4ax}$

$x^4=64b^{2}ax$

x = 0 or $x=4b^{2/3}{a}^{1/3}$

$A={\int }_0^{4b^{2/3/a^{2/3}}}\left(\sqrt{4ax}-\frac{x^2}{4b}\right)dx=\frac{16ab}{3}$

2. Whole area of the ellipse,  $x^2/a^2+y^2/b^2=1$ is $\pi ab$ sq. units.
3. Area included between the parabola $y^2=4ax$ & the line y = mx is $8a^2/3m^3$ sq. units.
4. The area of the region bounded by one arch of sin ax (or cos ax) and x-axis is $2/a$ sq. units. 
5. Average value of a function y = f(x) over an interval a $\le x\le b$ is defined as:

$y(av)=\frac{1}{b-a}{\int }_a^{b}f(x)dx$

6. If y = f(x) is a monotonic function in (a,b), then the area bounded by the ordinates at x = a, x= b, y = f(x) and y = f(c)$[where{c}_{\in }(a,b)]$ is minimum when $c=\frac{a+b}{2}$ .

6.0Solved Examples on Area Under the Curve

Example 1: Find area bounded by$x=1,x=2,y=x^2,y=0$ . 

Solution:

$A={\int }_1^2\left(x^2-0\right)dx$

$=\frac{{\left[x^3\right]}_1^2}{3}$

$=\frac{8-1}{3}=\frac{7}{3}$

Example 2: Find the area in the first quadrant bounded by $y=4x^2,x=0,y=1\text{ and }y=4$ .

Solution:

Required area = ${\int }_1^{4}xdy={\int }_1^4\frac{\sqrt{y}}{2}dy$

$=\frac{1}{2}\cdot \frac{2}{3}\cdot {\left(y^{\frac{3}{2}}\right)}_1^4$

$=\frac{1}{3}\left[4^{\frac{3}{2}}-1\right]$

$=\frac{1}{3}[8-1]=\frac{7}{3}=2\frac{1}{3}\text{ sq. units. }$

Example 3: Find the area enclosed between $y=\sin x;y=\cos x$ and y-axis in the 1^st quadrant

Solution:

$A={\int }_0^{\frac{\pi }{4}}(\cos x-\sin x)dx$

$=[\sin x+\cos x{]}_0^{\pi /4}=\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)-(0+1)$

$=\sqrt{2}-1$

Example 4: Find the area bounded by the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ 

Solution:

Area bounded by ellipse in first quadrant = ${\int }_0^3\frac{2}{3}\sqrt{9-x^2}dx=\frac{3\pi }{2}$

∵ Curve is symmetrical about all four quadrants

∴ Total area = 4 (Area in any one of the quadrants)

$=4\left(\frac{3\pi }{2}\right)=6\pi$

Example 5: Compute the area of the figure bounded by the parabolas $x=-2y^2,x=1-3y^2\text{. }$ 

Solution:

Solving the equations $x=-2y^2,x=1-3y^2$, 

we find that ordinates of the points of intersection 

of the two curves as $y_1=-1,y_2=1$

The points are (–2, –1) and (–2, 1).

The required area

$2{\int }_0^1\left(x_1-x_2\right)dy=2{\int }_0^1\left[\left(1-3y^2\right)-\left(-2y^2\right)\right]dy=2{\int }_0^1\left(1-y^2\right)dy$

$=2{\left[y-\frac{y^3}{3}\right]}_0^1=\frac{4}{3}\text{ sq.units. }$

7.0Practice Problems on Area Under the Curve

1. Find the area bounded by y = x² + 2 above  x-axis between x = 2 and x = 3.

2. Find the area bounded by the curve y = cos x  and the x-axis  from  x = 0 to x = 2π.

3. Find the area bounded by x = 2, x = 5, y = x², y = 0.

4. Find the area bounded by  y = x² and y = x.

5. A figure is bounded by the curvesy =\left|\sqrt{2} \sin \frac{\pi x}{4}\right|, y = 0, x = 2 and x = 4. At what angles to the positive x-axis straight lines must be drawn through (4, 0) so that these lines partition the figure into three parts of the same area.

8.0Solved Questions on Area Under the Curve

• How to Find Area Under the Curve?

Ans: The area under the curve is calculated using definite integrals. The integral of a function f(x) from a to b is given by:

$\text{ Area }={\int }_a^{b}f(x)dx$

This integral represents the net area between the curve and the x-axis over the interval [a, b].

• What is the Formula for the Area Under the Curve?

Ans: The formula to find the area under the curve is:

$\text{ Area }={\int }_a^{b}f(x)dx$

where f(x) is the function describing the curve, and a and b are the lower and upper limits of the interval.